Problem: What is the slope of the line tangent to $f(x) = 2x^{2}+4x-6$ at $x = -3$ ?
The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}+4(x+h)-6) - (2x^{2}+4x-6)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})+4(x+h)-6) - (2x^{2}+4x-6)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}+4x+4h-6-2x^{2}-4x+6}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}+4h}{h}$ $ = \lim_{h \to 0} 4x+2h+4$ $ = 4x+4$ $ = (4)(-3)+4$ $ = -8$